A solving method based on neural network for a class of multi-leader–follower games

Abstract

In this paper, we present a new solving approach for a class of multi-leader–follower games. For the problem studied, we firstly propose a neural network model. Then, based on Lyapunov and LaSalle theories, we prove that the trajectory of the neural network model can converge to the equilibrium point, which corresponds to the Nash equilibrium of the problem studied. The numerical results show that the proposed neural network approach is feasible to the problem studied.

Appendix

A. Proof of Proposition 3.1

Following Proposition 2 in [18], Proposition 3.1 can be obtained immediately. $\square$

B. Proof of Theorem 3.4

It is obvious that $E(z^{*})=0$ if and only if $z^{*}$ solves (8). By Theorem 3.3, $z^{*}$ satisfies (8) $\Leftrightarrow$ $x^{*}$ solves the variational inequality problem (7). Therefore, Theorem 3.4 is proved. $\square$

C. Proof of Theorem 3.5

Arbitrarily given $z^{*}\in \Theta$, By Proposition 3.1, $E(z^{*})=0$. It means that $z^{*}$ solves the unconstrained programming problem $\min \nolimits _{z\in R^{n_{i}+n_{ii}+s_{i}+s_{ii}+t_{i}+t_{ii}}}E(z)$. Following the necessary optimality conditions of optimization problem, we can deduce $\nabla E(z^{*})=0$. Then, $z^{*}\in \Omega$. As $z^{*}$ is an arbitrary point in the set $\Theta$. Then, $\Theta \subseteq \Omega$. $\square$

D. Proof of Theorem 3.6

If $F_{0}(x_{i},x_{ii})$ is strictly monotone, by Theorem 3.3 the system (8) has solutions. Moreover for $\forall z\ne z^{*}, E(z)>0$. It means that E(z) is positive and definitive. Let $z=z(t,z^{0})$ denote the trajectory of the neural network (10) corresponding to the initial point $z^{0}$, along it there is

$$\begin{aligned} \frac{{\text {d}}}{{\text {d}}t}E(z)=\, & {} \frac{{\text {d}}}{{\text {d}}t}E[z(t,z^{0})]\\=\, & {} \nabla E(z)^{T}\cdot \frac{{\text {d}}z}{{\text {d}}t}\\=\, & {} -k\Vert \nabla E(z)\Vert ^{2}<0 \end{aligned}$$

Following the Lyapunov Theorem [20], we have the result. $\square$

E. Proof of Theorem 3.7

Firstly, we prove the following two results.

• (a) The function $E(z(t,z^{0}))$ is monotone non-increasing with its variable.

  Actually, following Theorem 3.6, the result (a) is obvious.

• (b) Let $\Upsilon =\{z(t,z^{0}):t\ge 0\}$, then $z(t,z^{0})$ is a bound positive semi-trajectory.

  Firstly, E(z) is bound from below. Following the continuity of E(z) and with (a), we have that the set $L(z^{0})$ is compact and

  $$\begin{aligned} \Upsilon \subseteq L(z^{0}). \end{aligned}$$

  Hence, the result (b) is obtained.

Now, we will prove $\lim \nolimits _{n\rightarrow +\infty }z(t_{n},z^{0})={\bar{z}}$.

Firstly, $\Upsilon$ is a bound set of points. Take rigidly increasing time sequence ${\bar{t}}_{n}, 0\le {\bar{t}}_{1}<{\bar{t}}_{2}<\cdots <{\bar{t}}_{n}\rightarrow +\infty$, it is obvious that $\{z({\bar{t}}_{n},z^{0})\}$ is a bound sequence, which composes of infinite points. Then, there exists a trajectory $z(t_{n},z^{0})$ with the rigidly increasing time sequence $\{t_{n}\}\subseteq \{{\bar{t}}_{n}\}, t_{n}\rightarrow +\infty$, satisfying

$$\begin{aligned}&\lim \limits _{n\rightarrow +\infty }z(t_{n},z^{0})={\bar{z}}. \end{aligned}$$

In the following contents, we will prove the second part of the theorem.

Following the result in (a), we know that $E({\bar{z}})=0\Leftrightarrow \nabla E({\bar{z}})=0$. Following LaSalle invariance principle [21], there exists a set $\vartheta$ such that $\lim \nolimits _{t\rightarrow \infty }\Upsilon \rightarrow \vartheta$, where $\vartheta$ denotes the largest invariant set in the set of the equilibrium point. That means, there exists a time sequence $\{t_{n}\}$, satisfying

$$\begin{aligned}&\lim \limits _{n\rightarrow +\infty }z(t_{n},z^{0})={\bar{z}}, \end{aligned}$$

and ${\bar{z}}$ is the equilibrium point of the neural network (10). The proof is completed. $\square$

F. Proof of Theorem 3.8

Following Theorems 3.5, 3.6 and  3.7, Theorem 3.8 is obvious. $\square$